Mathematical Proof that Progressions will never Overcome a Negative Expectation Game
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Both are untrue statements. If you graph the bankroll, it will be a wildly oscillating line, going positive and negative, with increasing amplitude over time. It does not asymptote to a continuous net loss, unless a table limit was imposed.
Given enough time, you're guaranteed to win, per QFIT, and it only takes once.
The probability of going broke is exactly zero. The probability of being ahead at some point is greater than zero. Therefore, given enough trials, you're going to win. Eventually.
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The bias against progressions is understandable and justified (and I share it), but it's clouding many people's ability to think objectively about the question. Thorp was correct.
Given enough time, you're guaranteed to win, per QFIT, and it only takes once.
The probability of going broke is exactly zero. The probability of being ahead at some point is greater than zero. Therefore, given enough trials, you're going to win. Eventually.
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The bias against progressions is understandable and justified (and I share it), but it's clouding many people's ability to think objectively about the question. Thorp was correct.
Infinity is a long time. You may never get to your first win. At any point in infinity, you still have an infinity to go.
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You are caught in an endless series of denial. :laugh:
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Each time a martingale is restarted after a 1 unit win could be the beginning of a disastrous sequence where losing continues until the end of time. Chance of a disastrous sequence never happening given all the time in the world approaches 0 just like the chance of never winning until the end of time approaches 0. So which is prevalent? We have this hint: In a negative EV game the expectation is to lose an amount approaching infinity for the committed martingale player.
You still do not understand. Unlimited bankroll means that you can keep playing until you have a single win. At that point, you have a profit.
As for the graph, I'll point you to qfit's own:
http://www.blackjackincolor.com/useless4.htm
It's disastrous for any finite bankroll of course. But there is a clear upward trend, is there not?
As for the graph, I'll point you to qfit's own:
http://www.blackjackincolor.com/useless4.htm
It's disastrous for any finite bankroll of course. But there is a clear upward trend, is there not?
Go read your post again. You were the one who said the gun would always fire, given enough pulls. You've provided a nice analogy that exactly proves my point (and Thorp's). Are you admitting your analogy is incorrect? Will the gun ever fire, or not?
The only case that's being excluded is the one where no hand is ever won, for an infinite time. For this to be the case, the odds of winning would need to be identically zero. Because the odds of winning are nonzero, there is a guarantee of a win within an infinite time. You can call it "almost surely" if you like, but it's as close to certain as anything is.
Your attempt to lump me in with "progressionists" is nothing less than insulting. You know full well that I agree that progressions are completely worthless in any realistic situation. Once again, I'm arguing only that there is no proof here that Thorp's statement was debunked.
The only case that's being excluded is the one where no hand is ever won, for an infinite time. For this to be the case, the odds of winning would need to be identically zero. Because the odds of winning are nonzero, there is a guarantee of a win within an infinite time. You can call it "almost surely" if you like, but it's as close to certain as anything is.
Your attempt to lump me in with "progressionists" is nothing less than insulting. You know full well that I agree that progressions are completely worthless in any realistic situation. Once again, I'm arguing only that there is no proof here that Thorp's statement was debunked.
In a negative expectation game, martingale works only in the sense that you quit playing for good after making a profit. If winning one unit is your sole goal, any other system can do it because of variance in the short term.
However, if you keep playing the negative expectation game, your cumulative profit = (your disadvantage)*(total wager). It will never become plus. The more you play the more you lose.
Your cumulative profit is proportional to your total wager, but the slope is negative. How can you make the equation (cumulative profit) = -0.5%*(total wager) positive?
However, if you keep playing the negative expectation game, your cumulative profit = (your disadvantage)*(total wager). It will never become plus. The more you play the more you lose.
Your cumulative profit is proportional to your total wager, but the slope is negative. How can you make the equation (cumulative profit) = -0.5%*(total wager) positive?
Other systems (or finite table limits) "can" do it if you're lucky. But over time, the bankroll function always decreases and asymptotes to a pure loss of slope -EV.
With martingale, like I said, the bankroll is a diverging, oscillating function that grows with magnitude over time. If you keep playing, you'll always go positive and negative at various times. Pick a positive time, which, per QFIT's example, will always arrive, and you're a winner.
Your equation describes the expectation at any specific number of hands (or total wager), but because the function is so unstable, it's a poor predictor.
What's the sum of the series 1-2+3-4...? But what are the values you'll actually see? It all depends on when you decide to stop.
With martingale, like I said, the bankroll is a diverging, oscillating function that grows with magnitude over time. If you keep playing, you'll always go positive and negative at various times. Pick a positive time, which, per QFIT's example, will always arrive, and you're a winner.
Your equation describes the expectation at any specific number of hands (or total wager), but because the function is so unstable, it's a poor predictor.
What's the sum of the series 1-2+3-4...? But what are the values you'll actually see? It all depends on when you decide to stop.
Your cumulative profit is only whatever you made at the point you decide to stop. Since you get to choose to stop where you want, you can claim whatever profit you wish.
Can you prove to me it's always negative? Or never positive? Look at the series example I just gave and think about it carefully.
You can't make it positive for all or any arbitrary future point; you have to end on a point where there is a profit. Because the series oscillates, this point will occur ('almost certainly'!). This does not happen when there is a table limit, because eventually the oscillation can't grow large enough to maintain any positive points. THEN the series asymptotes to -0.5*(wager).
Can you prove to me it's always negative? Or never positive? Look at the series example I just gave and think about it carefully.
You can't make it positive for all or any arbitrary future point; you have to end on a point where there is a profit. Because the series oscillates, this point will occur ('almost certainly'!). This does not happen when there is a table limit, because eventually the oscillation can't grow large enough to maintain any positive points. THEN the series asymptotes to -0.5*(wager).
QFIT said:
According to Euler and later shown by Niels Abel, the sum of 1-2+3-4 taken to infinity is 1/4. The general case is:
In your example, x=1 and the sum is 1/4.
In your example, x=1 and the sum is 1/4.
No, you get an oscillation from positive to negative values that grows in amplitude indefinitely. Just like in a martingale. That's why stating it purely in terms of EV is a poor reflection of what's going on mathematically.